[proofplan] The plurisubharmonicity of $\phi$ is, by definition, the requirement that every complex-line restriction be subharmonic in the one-variable sense. We define the slice function $u: D(0, R) \to [-\infty, \infty)$ by $u(\zeta) = \phi(a + \zeta v)$ on an open disc strictly containing $\overline{D}(0,r)$. By the slice clause of the definition of psh, $u$ is subharmonic. The circular sub-[mean value inequality](/theorems/328) for subharmonic functions, applied to $u$ at the origin with radius $r$, yields the stated estimate. The degenerate case $v = 0$ reduces the right-hand side to $\phi(a)$ and the inequality holds with equality. [/proofplan] [step:Reduce to the degenerate case $v = 0$ and dispose of it] If $v = 0$, then for every $\theta \in [0, 2\pi]$ we have $a + r e^{i\theta} v = a$, hence \begin{align*} \frac{1}{2\pi} \int_0^{2\pi} \phi(a + r e^{i\theta} v) \, d\mathcal{L}^1(\theta) = \frac{1}{2\pi} \int_0^{2\pi} \phi(a) \, d\mathcal{L}^1(\theta) = \phi(a), \end{align*} so the inequality holds with equality (the integrand is $\mathcal{L}^1$-measurable as a constant function with value in $[-\infty, \infty)$, and the integral is interpreted in the extended sense if $\phi(a) = -\infty$). We may therefore assume $v \ne 0$ for the remainder of the proof. [guided] The case $v = 0$ is a notational degeneracy: the complex line $\{a + \zeta v : \zeta \in \mathbb{C}\}$ collapses to the single point $\{a\}$, and the "slice" $u(\zeta) = \phi(a + \zeta v) = \phi(a)$ is constant. The hypothesis $\overline{D}(0,r) \subseteq \{\zeta : a + \zeta v \in \Omega\}$ is then trivially satisfied because $a \in \Omega$ implies every $\zeta$ maps to $a \in \Omega$. The circular average of the constant function $\phi(a)$ on $[0, 2\pi]$ is $\phi(a)$, so both sides of the asserted inequality equal $\phi(a)$. We discard this case and assume $v \ne 0$, which is what is needed to identify the slice as a genuine complex affine line. [/guided] [/step] [step:Construct the slice function on an open neighbourhood of $\overline{D}(0, r)$] Define the affine map \begin{align*} L: \mathbb{C} &\to \mathbb{C}^n \\ \zeta &\mapsto a + \zeta v, \end{align*} and the preimage \begin{align*} V := L^{-1}(\Omega) = \{\zeta \in \mathbb{C} : a + \zeta v \in \Omega\}. \end{align*} Since $L$ is continuous and $\Omega \subseteq \mathbb{C}^n$ is open, $V \subseteq \mathbb{C}$ is open. By hypothesis $\overline{D}(0, r) \subseteq V$, and $\overline{D}(0, r)$ is compact while $V$ is open, so there exists $R > r$ with $\overline{D}(0, r) \subseteq D(0, R) \subseteq V$ (take $R := r + \tfrac{1}{2} \operatorname{dist}(\overline{D}(0, r), \mathbb{C} \setminus V)$, finite and positive because the complement $\mathbb{C} \setminus V$ is either empty or a [closed set](/page/Closed%20Set) disjoint from the compact $\overline{D}(0, r)$; in the empty case set $R := r + 1$). Define the slice \begin{align*} u: D(0, R) &\to [-\infty, \infty) \\ \zeta &\mapsto \phi(a + \zeta v) = \phi(L(\zeta)). \end{align*} [guided] Why do we bother enlarging the disc from $\overline{D}(0,r)$ to an open disc $D(0,R)$ with $R > r$? Because subharmonicity is a property of functions on **open** sets: we need $u$ to be defined and subharmonic on an open neighbourhood of $\overline{D}(0,r)$ in order to apply the circular sub-[mean value inequality](/theorems/328) at radius $r$. The compactness-openness argument is standard: a compact set inside an [open set](/page/Open%20Set) has positive distance to the complement (or the complement is empty, in which case $V = \mathbb{C}$ and any $R$ works). Concretely, if $V = \mathbb{C}$ — which happens, for instance, when $\Omega = \mathbb{C}^n$ — we just pick $R = r + 1$. Otherwise $d := \operatorname{dist}(\overline{D}(0,r), \mathbb{C} \setminus V) > 0$ because the distance from a compact set to a disjoint [closed set](/page/Closed%20Set) is strictly positive, and we set $R = r + d/2$. The slice $u$ is well-defined on $D(0,R)$ because $L(D(0,R)) \subseteq L(V) \subseteq \Omega$, the domain of $\phi$. It takes values in $[-\infty, \infty)$ since $\phi$ does. [/guided] [/step] [step:Invoke plurisubharmonicity to conclude $u$ is subharmonic on $D(0, R)$] By the definition of plurisubharmonicity, $\phi: \Omega \to [-\infty, \infty)$ is upper semicontinuous, not identically $-\infty$ on any connected component of $\Omega$, and for every $b \in \Omega$ and every $w \in \mathbb{C}^n$, the slice function \begin{align*} \zeta \mapsto \phi(b + \zeta w) \end{align*} is subharmonic on each connected component of $\{\zeta \in \mathbb{C} : b + \zeta w \in \Omega\}$. Applying this with $b = a$ and $w = v$, the function $u(\zeta) = \phi(a + \zeta v)$ is subharmonic on each connected component of $V$, hence in particular on the connected open subset $D(0, R) \subseteq V$. [guided] This step consumes the entire hypothesis "$\phi$ is psh". Plurisubharmonicity is **defined** as a slice-by-slice condition: $\phi$ is psh on $\Omega$ if and only if (i) it is upper semicontinuous, (ii) it is not identically $-\infty$ on any connected component, and (iii) every complex-line restriction $\zeta \mapsto \phi(b + \zeta w)$ is subharmonic on the [open set](/page/Open%20Set) $\{b + \zeta w \in \Omega\}$. We are using clause (iii) with $b = a$ and $w = v$. The resulting one-variable function $u$ is subharmonic on the (possibly disconnected) [open set](/page/Open%20Set) $V$. Since $D(0,R)$ is a connected open subset of $V$ — it is connected because it is convex — $u$ is subharmonic on $D(0,R)$ as well, by the standard fact that subharmonicity restricts to open subsets. Note that we do **not** need to know anything about $\phi$ off the slice $L(D(0,R))$; the entire argument lives on this one-dimensional disc. [/guided] [/step] [step:Apply the one-variable circular sub-mean value inequality at the origin] Since $u$ is subharmonic on the open disc $D(0, R)$ and $0 \in D(0, R)$ with $\overline{D}(0, r) \subseteq D(0, R)$, the circular sub-[mean value inequality](/theorems/328) for subharmonic functions yields \begin{align*} u(0) \le \frac{1}{2\pi} \int_0^{2\pi} u(r e^{i\theta}) \, d\mathcal{L}^1(\theta). \end{align*} Substituting the definition of $u$, \begin{align*} u(0) &= \phi(a + 0 \cdot v) = \phi(a), \\ u(r e^{i\theta}) &= \phi(a + r e^{i\theta} v), \end{align*} gives \begin{align*} \phi(a) \le \frac{1}{2\pi} \int_0^{2\pi} \phi(a + r e^{i\theta} v) \, d\mathcal{L}^1(\theta), \end{align*} which is the asserted inequality. (Citing a result not yet in the wiki: Sub-[Mean Value Inequality](/theorems/328) for Subharmonic Functions — the property that for any subharmonic function $u$ on an [open set](/page/Open%20Set) $U \subseteq \mathbb{C}$, any $\zeta_0 \in U$, and any $\rho > 0$ with $\overline{D}(\zeta_0, \rho) \subseteq U$, one has $u(\zeta_0) \le \frac{1}{2\pi} \int_0^{2\pi} u(\zeta_0 + \rho e^{i\theta}) \, d\mathcal{L}^1(\theta)$. This is part of the standard definition or characterisation of subharmonicity.) [guided] We apply the circular sub-[mean value inequality](/theorems/328) for subharmonic functions to $u$ at the centre $\zeta_0 = 0$ with radius $\rho = r$. The hypotheses are: $u$ is subharmonic on an [open set](/page/Open%20Set) $U = D(0, R)$ (verified in the previous step); $0 \in U$ (immediate, $|0| = 0 < R$); and $\overline{D}(0, r) \subseteq U = D(0, R)$ (immediate, $r < R$ by construction). The integrand $\theta \mapsto u(r e^{i\theta}) = \phi(a + r e^{i\theta} v)$ is $\mathcal{L}^1$-measurable on $[0, 2\pi]$ because $\phi$ is upper semicontinuous (and hence Borel measurable) and $\theta \mapsto a + r e^{i\theta} v$ is continuous; the integral is well-defined in $[-\infty, \infty)$ because $\phi$ is bounded above on the compact circle $\{a + r e^{i\theta} v : \theta \in [0, 2\pi]\} \subseteq \Omega$ by upper semicontinuity. The substitution $u(0) = \phi(a)$ and $u(re^{i\theta}) = \phi(a + re^{i\theta} v)$ is just unwinding the definition of $u$; no change of variables in the integral is involved because $\theta$ remains the integration variable. This delivers exactly the asserted inequality and completes the proof. [/guided] [/step]